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In a certain game, a large container is filled with red, [#permalink]
19 Jun 2003, 13:24

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80% (02:20) correct
20% (00:11) wrong based on 31 sessions

In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

Re: Hardest GMAT Problem Solving I've Seen [#permalink]
19 Jun 2003, 17:04

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This post received KUDOS

Curly05 wrote:

. In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

I get 2.

First I simplified 147,000 to 147.
147 divides by 3, since 1+4+7 is divisible by 3.
147 divided by 3 = 49, which in turn equals 7x7.

So 7x7x3 = 147.

Now we need to get 147 to equal 147,000. We do this by multiplying 147 x 1000.
To get 1000, multiply (5x2)x(5x2)x(5x2).

Re: Hardest GMAT Problem Solving I've Seen [#permalink]
06 Oct 2009, 10:22

Expert's post

JP, you make it so easy to understand. I'm beginning to realize now what people have been telling me on these forums: the math uses basic concepts. The more difficult problems are just dressed up in something scary.

But I still freak out sometimes. _________________

Re: Hardest GMAT Problem Solving I've Seen [#permalink]
17 Jun 2010, 10:55

I think the easiest way of thinking about it is factor out seven as many times as you can (each time you factor out a seven you're counting a red ball). 147000/7=21000, 21000/7=3000, and it's pretty easy to see (or check) that 3000 is some multiple of 2,3, and 5.

Re: Hardest GMAT Problem Solving I've Seen [#permalink]
05 Jan 2011, 17:47

You psyched yourself out. A question like this can and should be easily done in under a minute. A light bulb should turn on as you think...What's the concept here...AHA! FACTORIZATION! If that light bulb doesn't go off, you are not prepared enough, because virtually every GMAT problem has a label.

Re: Hardest GMAT Problem Solving I've Seen [#permalink]
05 Jan 2011, 21:10

I've seen this problem in some book.. I think MGMAT number properties.. And I've seen another versions of the same concept elsewhere too.

BTW yes, its been very nicely explained above. The key here is indeed understanding that the problem is just calling for prime factorization. _________________

Re: Hardest GMAT Problem Solving I've Seen [#permalink]
13 Aug 2011, 18:40

Curly05 wrote:

. In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

= 147,000 = 147 x 1000 = (3) x (7x7) x (2x2x2) x (5x5x5) = 2^3 x 3^1 x 5^3 x 7^2

Re: Hardest GMAT Problem Solving I've Seen [#permalink]
17 Aug 2011, 10:11

Is it just me or are there no answer choices posted? Without looking at the answer choices you can't determine the solution. For example you could remove 21,000 red beads and no others, which would give a total of 147,000. If this were on the GMAT I think they would give a limit to the number of beads of each color or make it obvious from the answer choices how to approach it.

Re: Hardest GMAT Problem Solving I've Seen [#permalink]
17 Aug 2011, 11:21

mj12g wrote:

Is it just me or are there no answer choices posted? Without looking at the answer choices you can't determine the solution. For example you could remove 21,000 red beads and no others, which would give a total of 147,000. If this were on the GMAT I think they would give a limit to the number of beads of each color or make it obvious from the answer choices how to approach it.

There is only one answer per the given conditions. No need of options.

We can definitely say that only 2 red beads were removed, no more no less. We can deduce that by seeing the exponent of 7, which is 2.

Re: In a certain game, a large container is filled with red, [#permalink]
05 Apr 2012, 13:19

fluke wrote:

mj12g wrote:

Is it just me or are there no answer choices posted? Without looking at the answer choices you can't determine the solution. For example you could remove 21,000 red beads and no others, which would give a total of 147,000. If this were on the GMAT I think they would give a limit to the number of beads of each color or make it obvious from the answer choices how to approach it.

There is only one answer per the given conditions. No need of options.

We can definitely say that only 2 red beads were removed, no more no less. We can deduce that by seeing the exponent of 7, which is 2.

147000=2^3*3*5^3*7^2

Ans: "2"

Right, this is because the question stem says 'product' of the point values. In the example where we remove 21,000 red beads, the total must be calculated as the product of the point values for 21,000 beads which is 7^21,000. A truly massive number!

Re: In a certain game, a large container is filled with red, [#permalink]
05 Apr 2012, 13:29

1

This post received KUDOS

Expert's post

Curly05 wrote:

In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed?

Re: In a certain game, a large container is filled with red, [#permalink]
01 May 2013, 02:37

Not sure if Red = 2 is the only answer. Other answers: Red = 7 (with Y, G, B as 4, 5 , 5) Red = 5 ( with Y, G, B as 7, 2, 10) Red = 4 (with Y, G, B as 5, 5, 7) Red = 2 (with Y, G, B as 10, 5, 7) (values of Y, G, B in any sequence)

And if Red = 2, then by default, is Red = 4 not also possible?

What if no Reds have been taken out? Surely Red = 0 is also possible?

Re: In a certain game, a large container is filled with red, [#permalink]
01 May 2013, 07:36

Expert's post

EdNyhof wrote:

Not sure if Red = 2 is the only answer. Other answers: Red = 7 (with Y, G, B as 4, 5 , 5) Red = 5 ( with Y, G, B as 7, 2, 10) Red = 4 (with Y, G, B as 5, 5, 7) Red = 2 (with Y, G, B as 10, 5, 7) (values of Y, G, B in any sequence) And if Red = 2, then by default, is Red = 4 not also possible?

What if no Reds have been taken out? Surely Red = 0 is also possible?

Let me know what you think.

Is not clear what you mean by that.The problem states "product of the point values of the removed beads is 147,000."All the given points (2,3,5,7) are primes. The given point value is 147,000 = 49*30*100 = 7^2*5^3*2^3*3. Now if I would have removed one yellow bead and one blue bead, the product of the points would be = 5*2. Thus, in no other way can you get a factor of 7, if you don't pick up 2 red beads. _________________

Re: In a certain game, a large container is filled with red, [#permalink]
19 Sep 2013, 17:32

mau5 wrote:

EdNyhof wrote:

Not sure if Red = 2 is the only answer. Other answers: Red = 7 (with Y, G, B as 4, 5 , 5) Red = 5 ( with Y, G, B as 7, 2, 10) Red = 4 (with Y, G, B as 5, 5, 7) Red = 2 (with Y, G, B as 10, 5, 7) (values of Y, G, B in any sequence) And if Red = 2, then by default, is Red = 4 not also possible?

What if no Reds have been taken out? Surely Red = 0 is also possible?

Let me know what you think.

Is not clear what you mean by that.The problem states "product of the point values of the removed beads is 147,000."All the given points (2,3,5,7) are primes. The given point value is 147,000 = 49*30*100 = 7^2*5^3*2^3*3. Now if I would have removed one yellow bead and one blue bead, the product of the points would be = 5*2. Thus, in no other way can you get a factor of 7, if you don't pick up 2 red beads.

Could you please explain how you got this: [color=#ff0000]7^2*5^3*2^3*3[/color] ?

Re: In a certain game, a large container is filled with red, [#permalink]
20 Sep 2013, 00:50

1

This post received KUDOS

Expert's post

azizf60 wrote:

mau5 wrote:

EdNyhof wrote:

Not sure if Red = 2 is the only answer. Other answers: Red = 7 (with Y, G, B as 4, 5 , 5) Red = 5 ( with Y, G, B as 7, 2, 10) Red = 4 (with Y, G, B as 5, 5, 7) Red = 2 (with Y, G, B as 10, 5, 7) (values of Y, G, B in any sequence) And if Red = 2, then by default, is Red = 4 not also possible?

What if no Reds have been taken out? Surely Red = 0 is also possible?

Let me know what you think.

Is not clear what you mean by that.The problem states "product of the point values of the removed beads is 147,000."All the given points (2,3,5,7) are primes. The given point value is 147,000 = 49*30*100 = 7^2*5^3*2^3*3. Now if I would have removed one yellow bead and one blue bead, the product of the points would be = 5*2. Thus, in no other way can you get a factor of 7, if you don't pick up 2 red beads.

Could you please explain how you got this: [color=#ff0000]7^2*5^3*2^3*3[/color] ?

2^3*3*5^3*7^2is prime factorization of 147,000 = 1,000*147 = (8*125)*(3*49) = 2^3*3*5^3*7^2. _________________